How is the option price of an plain vanilla option (in a Black Scholes setting) derived, which is written on, say XAGGBP but practically hedged with XAGUSD and GBPUSD (because these are more liquid)? Eventually, I am interested in the delta(s) and correlation risk with respect to XAGUSD and GBPUSD.
Answer
Let Xxag→gbpt be the exchange rate from one unit of XAG to units of GBP. Moreover, let Xxag→usdt, and Xgbp→usdt be the respective exchanges rates from one unit of XAG and GBP to units of USD. We consider an option payoff, in GBP, at maturity T of the form (Xxag→gbpT−K)+. Note that (Xxag→gbpT−K)+=(Xxag→usdTXgbp→usdT−K)+. We assume that, under the USD risk-neutral probability measure Qusd, dXxag→usdt=Xxag→usdt[(rusd−rxag)dt+σ1dW1t],dXgbp→usdt=Xgbp→usdt[(rusd−rgbp)dt+σ2(ρdW1t+√1−ρ2dW2t)], where rusd, rgbp, and rxag are interest rates, σ1 and σ2 are volatilities, ρ is the correlation, and {W1t,t≥0} and {W2t,t≥0} are two standard independent Brownian motions.
Let Busdt=erusdt and Bgbpt=ergbpt be the respective USD and GBP money market account values at time t. Moreover, let Qgbp be the GBP risk-neutral probability measure. Note that dQgbpdQusd|t=BgbptXgbp→USDtBusdtXgbp→USD0=e−12σ22t+σ2(ρW1t+√1−ρ2W2t). Then, {˜W1t,t≥0} and \{\tilde{W}_t^2, \, t\ge 0\}, where \begin{align*} \tilde{W}_t^1 &= W_t^1 - \sigma_2\rho t, \\ \tilde{W}_t^2 &= W_t^2 - \sigma_2\sqrt{1-\rho^2} t, \end{align*} are two standard independent Brownian motions under Q_{gbp}. Furthermore, under Q^{gbp}, \begin{align*} dX_t^{xag\rightarrow usd} &= X_t^{xag\rightarrow usd}\left[\left(r^{usd}-r^{xag} +\rho\sigma_1\sigma_2\right)dt +\sigma_1 d\tilde{W}_t^1 \right],\\ dX_t^{gbp\rightarrow usd} &= X_t^{gbp\rightarrow usd}\left[\left(r^{usd}-r^{gbp} +\sigma_2^2\right)dt +\sigma_2\left(\rho d\tilde{W}_t^1 +\sqrt{1-\rho^2}d\tilde{W}_t^2\right)\right]. \end{align*} Then \begin{align*} X_t^{xag\rightarrow gbp} &= \frac{X_t^{xag\rightarrow usd}}{X_t^{gbp\rightarrow usd}} \\ &=\frac{X_0^{xag\rightarrow usd}}{X_0^{gbp\rightarrow usd}} e^{\left(r^{gbp}-r^{xag}+\rho\sigma_1\sigma_2-\frac{1}{2}\sigma_1^2 -\frac{1}{2}\sigma_2^2\right)t + (\sigma_1-\rho\sigma_2)\tilde{W}_t^1 -\sigma_2\sqrt{1-\rho^2}\tilde{W}_t^2}\\ &=X_0^{xag\rightarrow gbp} e^{\left(r^{gbp}-r^{xag}-\frac{\sigma_1^2+\sigma_2^2-2\rho\sigma_1\sigma_2}{2}\right)t + \sqrt{\sigma_1^2+\sigma_2^2 - 2\rho\sigma_1\sigma_2 }\frac{(\sigma_1-\rho\sigma_2)\tilde{W}_t^1 -\sigma_2\sqrt{1-\rho^2}\tilde{W}_t^2}{\sqrt{\sigma_1^2+\sigma_2^2 - 2\rho\sigma_1\sigma_2 }} }. \end{align*}
Let \begin{align*} \sigma = \sqrt{\sigma_1^2+\sigma_2^2 - 2\rho\sigma_1\sigma_2 }, \end{align*} and \begin{align*} W_t^3 = \frac{(\sigma_1-\rho\sigma_2)\tilde{W}_t^1 -\sigma_2\sqrt{1-\rho^2}\tilde{W}_t^2}{\sqrt{\sigma_1^2+\sigma_2^2 - 2\rho\sigma_1\sigma_2 }}. \end{align*} Then \{W_t^3, \, t \ge 0\} is a standard Brownian motion under Q^{gbp}, by Levy's characterization. Moreover, \begin{align*} d X_t^{xag\rightarrow gbp} = X_t^{xag\rightarrow gbp}\left[\left(r^{gbp}-r^{xag} \right)dt +\sigma dW_t^3 \right]. \end{align*} Therefore, the option payoff (1) can be valued using the Garman Kohlhagen formula, while replace the initial exchange rate X_0^{xag\rightarrow gbp} by \frac{X_0^{xag\rightarrow usd}}{X_0^{gbp\rightarrow usd}}. The respective hedge ratios can be computed subsequently.
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